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3 years ago

How does calcium obey the octet rule when reacting to form compounds?

Chemistry

2 answers:

wolverine [178]3 years ago

8 0

The answer is C. because it gives up two electrons.

Leya [2.2K]3 years ago

5 0

Answer: Option (C) is the correct answer.

Explanation:

Atomic number of calcium is 20 and its electronic distribution is 2, 8, 8, 2.

Therefore, a calcium atom contains two valence electrons and in order to attain stability it needs to lose two electrons.

Hence, it acquires a charge of +2 as $Ca^{2+}$ by loss of two electrons.

As, according to the octet rule there must be eight electrons in the valence shell.

Thus, we can conclude that calcium obey the octet rule when reacting to form compounds as gives up electrons.

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The human body can get energy by metabolizing proteins, carbohydrates or fatty acids, depending on the circumstances. Roughly sp

Kamila [148]

Answer:

the ratio of the energy the body gets metabolizing each gram of alanine to the energy the body gets metabolizing each gram of glucose is 1.0111

Explanation:

Given the data in the question;

To determine the ratio of the energy the body gets metabolizing each gram of alanine to the energy the body gets metabolizing each gram of glucose, first we get the molar masses of both alanine and glucose

we know that;

Molar mass of alanine ( C₃H₇NO₂ ) = 89.09 g/mol

Molar mass of glucose ( C₆H₁₂O₆ ) = 180.16 g/mol

now, { metabolizing each gram }

moles of alanine = mass taken / molar mass

= 1g / 89.09 g/mol = 1/89.09 moles

moles of glucose = mass taken / molar mass

= 1g / 180.16 g/mol = 1/180.16 moles

In each molecule of alanine, we have 3 atoms of carbon.

Also, in each molecules of glucose, we have 6 atoms of carbon

so,

number of moles of Carbons in alanine = 3 × 1/89.09 moles = 0.03367

number of moles of Carbons in glucose = 6 × 1/180.16 moles = 0.0333

so ratio of energy will be the ratio of carbon atoms, which is;

⇒ 0.03367 / 0.0333 = 1.0111

Therefore, the ratio of the energy the body gets metabolizing each gram of alanine to the energy the body gets metabolizing each gram of glucose is 1.0111

7 0

3 years ago

HELP PLEASE! :(

BabaBlast [244]

Answer:

Part 1: W = 116 Y = 163

Part 2: Since 232 is the mailing point of 2 kg then you would divide 232 by 2 to get the melting point for 1 kg, the same with Y.

6 0

3 years ago

Read 2 more answers

Why is it so hard to dispose of nuclear waste

Anarel [89]

It is highly hazardous, it is radioactive

8 0

3 years ago

If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is Group of answer choices

BaLLatris [955]

Answer:

The water is completely vaporized at this stage.

Explanation:

The complete question is

If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is

-boiling

-completely vaporized

-frozen solid

-decomposed

-still a liquid

Energy added = 50 kJ = 50000 J

mass of water = 15.5 g = 0.0155 kg

temperature of water = 10 °C

We know that the energy posses by a mass of water at a given temperature is given as

H = mcT

where H is the energy possessed by the mass of water

m is the mass of the water

c is the specific heat capacity of water = 4200 J/ kg- °C

T is the temperature of the water

substituting values, the energy of this amount of water is

H = 0.0155 x 4200 x 10 = 651 J

If 50 kJ is added to the water, the energy increases to

50000 J + 651 J = 50651 J

Temperature of this water at this stage will be gotten from

H = mcT

we solve for the new temperature

50651 = 0.0155 x 4200 x T

50651 = 65.1 x T

T = 50651/65.1 = 778.05 °C

This temperature is well over 100 °C, which is the vaporization temperature of water, but less than 3000 °C for its molecules to decompose.

3 0

3 years ago

A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal

goblinko [34]

Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂:

$1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC$

All of the hydrogens atoms in the compound ended up becoming H₂O, so the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O:

$0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH$

Because the compound is composed only by C, H and O, the mass of O in the compound can be calculated by substraction:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C

mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H

mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

8 0

3 years ago